Bayes' Theorem States : *If $E_1,E_2,….E_n$ are n non empty evnents which constitute a partition of sample space S, ie.e. $E_1,E_2,….E_n$ are pairwise disjoint and $E_1 \cup E_2 ……\cup E_n$ = S and A is any event of non zero probability , then $ P(E_i |A) = \frac{P(E_i)P(A|E_i)}{\sum^n_{j=1} P(E_j) P(A | E_j)}$ for any i = 1,2,3,…….n.*
Problem :
Suppose that the reliability of a HIV test is specified as follows : Of people having HIV, 90% of the test detect the disease but 10% go undetected. Of people free of HIV, 99% of the test judged HIV -ve but 1% are diagnosed as showing HIV +ve . From a large population of which only 0.1% have HIV, one person is selected at random , given the HIV test, and the pathologist reports him / her as HIV +ve. What is the probability that the person actually has HIV ?
Request you to please help how to proceed in such problem by using Bayes' theorem.. Thanks in advance
Best Answer
Let $A$ be the event "tested positive." Let $E_1$ be the event "has HIV" and $E_2$ the event "doesn't have HIV."
We want $\Pr(E_1|A)$.
On the top of the right-hand side of the formula you quoted, we have $\Pr(E_1)\Pr(A|E_1)$.
We are told that $\Pr(E_1)=0.001$. We are also told $\Pr(A|E_1)= 0.90$. So now we know the top.
For the bottom, we know $\Pr(E_1)\Pr(A|E_1)$, since we just have calculated it. We want $\Pr(E_2)\Pr(A|E_2)$. By the information we were given, the product is $(0.999)(0.01)$.
Now you have all the ingredients.